\(M=\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{xy}\)

\(=\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}+\frac{3}{4}.\frac{x^2+y^2}{xy}\)

\(\ge2\sqrt{\frac{xy}{x^2+y^2}.\frac{x^2+y^2}{4xy}}+\frac{3}{4}.\frac{2xy}{xy}\)

\(\Rightarrow M\ge1+\frac{3}{2}=\frac{5}{2}\)

Dấu = xảy ra khi \(x=y>0\)

Aps dụng bđt coossi rồi tách ghepos nha bạn

v cả quốc béo

Ta có: \(Q=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}=\dfrac{2}{x^2+y^2}+\dfrac{6}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}\)

Áp dụng BĐT phụ: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow2\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)\ge2\left(\dfrac{4}{x^2+2xy+y^2}\right)=2\left[\dfrac{4}{\left(x+y\right)^2}\right]=2.\dfrac{4}{4}=2\)

Dấu "=" xảy ra khi x=y=1

Áp dụng BĐT phụ: \(ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2^2}{4}=1\)

Dấu"=" xảy ra khi x=y=1

\(\Rightarrow2xy\le2.1=2\)

\(\Rightarrow\dfrac{4}{2xy}\ge\dfrac{4}{2}=2\)

\(\Rightarrow Q=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}\ge2+2=4\)

Dấu"=" xảy ra khi x=y=1

\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)

\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)

\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)

Áp dụng BĐT Schwarz : \(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}=4\)

Lại có \(\dfrac{1}{2xy}=\dfrac{2}{4xy}\ge\dfrac{2}{\left(x+y\right)^2}=2\)

Cộng vế với vế được P \(\ge6\) ("=" khi x = y = 1/2)

Vậy Min P = 6 <=> x = y = 1/2 

\(\frac{4x + y}{xy} = \frac{4x}{xy} + \frac{y}{xy} = \frac{4}{y} + \frac{1}{x}\)

\(P = \frac{1}{x} + \frac{4}{y} + \frac{2x - y}{4}\)

\(=\frac{1}{x}+\frac{4}{y}+\frac{x}{2}-\frac{y}{4}\)

Theo BĐT Cauchy, ta có:

\(\frac{1}{x} + \frac{x}{4} \ge 2\sqrt{\frac{1}{x} \cdot \frac{x}{4}} = 2 \cdot \frac{1}{2} = 1\)

\(\frac{4}{y} + \frac{4y}{9} \ge 2\sqrt{\frac{4}{y} \cdot \frac{4y}{9}} = 2 \cdot \frac{4}{3} = \frac{8}{3}\)

\(P = \left( \frac{1}{x} + \frac{x}{4} \right) + \left( \frac{4}{y} + \frac{4y}{9} \right) + \left( \frac{x}{2} - \frac{x}{4} \right) - \frac{y}{4} - \frac{4y}{9}\)

\(=\left(\frac{1}{x}+\frac{x}{4}\right)+\left(\frac{4}{y}+\frac{4y}{9}\right)+\frac{x}{4}-\frac{25y}{36}\)

x+y=5

=>y=5-x

\(\frac{x}{4} - \frac{25y}{36} = \frac{5 - y}{4} - \frac{25y}{36} = \frac{5}{4} - \frac{y}{4} - \frac{25y}{36} = \frac{5}{4} - \frac{34y}{36} = \frac{5}{4} - \frac{17y}{18}\)

=>\(P \ge 1 + \frac{8}{3} + \frac{5}{4} - \frac{17y}{18}\)

Chọn điểm rơi tại y=3

Khi đó, ta sẽ có: \(P\ge1+\frac{8}{3}+\frac{5}{4}-\frac{17 \cdot3}{18}=\frac{25}{12}\)

Dấu '=' xảy ra khi y=3 và x=5-y=5-3=2

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).

Đẳng thức xảy ra khi x = 1; y = 2.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)